Grassmann Algebra

TheRegressiveProduct.nb 32
Provided that the simplicity conditions on the coefficients are satisfied:
a1�a6 � a3 a4 � a2 a5 � 0
In sum: A 2-element in a 4-space may be factorized if and only if a condition on the
coefficients is satisfied.
a1�e1 � e2 � a2�e1 � e3 � a3�e1 � e4 �
a4�e2 � e3 � a5�e2 � e4 � a6�e3 � e4
� 1
������� ��a1 e1 � a4 e3 � a5 e4���a1 e2 � a2 e3 � a3 e4�,
a1
� a3 a4 � a2 a5 � a1 a6 � 0
Example 2: Factorizing a 3-element in a 5-space
This time we have a numerical example of a 3-element in 5-space. We wish to determine if the
element is simple, and if so, to obtain a factorization of it. To achieve this, we first assume that
it is simple, and apply the factorization algorithm. We will verify its simplicity (or its nonsimplicity)
by comparing the results of this process to the original element.
Α 3 ��3e1 � e2 � e3 � 4e1 � e2 � e4 �
12 e1 � e2 � e5 � 3e1 � e3 � e4 � 3e1 � e3 � e5 �
8e1 � e4 � e5 � 6e2 � e3 � e4 � 18 e2 � e3 � e5 � 12 e3 � e4 � e5
¥ Select a 3-element belonging to at least one of the terms, say e1 � e2 � e3 .
¥ Drop e1 to create e2 � e3 . Drop e2 to create e1 � e3 . Drop e3 to create e1 � e2 .
¥ Select e2 � e3 .
¥ Drop the terms not containing it, factor it from the remainder, and eliminate it to give Α1 .
Α1 ��3e1 � 6e4 � 18 e5 ��e1 � 2e4 � 6e5
¥ Select e1 � e3 .
¥ Drop the terms not containing it, factor it from the remainder, and eliminate it to give Α2 .
Α2 � 3e2 � 3e4 � 3e5 � e2 � e4 � e5
¥ Select e1 � e2 .
¥ Drop the terms not containing it, factor it from the remainder, and eliminate it to give Α3 .
Α3 ��3e3 � 4e4 � 12 e5
¥ The exterior product of these 1-element factors is:
3.38
Α1 � Α2 � Α3 � �� e1 � 2e4 � 6e5���e2 � e4 � e5����3 e3 � 4e4 � 12 e5�
� Multiplying this out (here using GrassmannSimplify) gives:
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